ExampleConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (claude-sonnet-5 + deepseek-v4-pro)verified 2026-08-05 (claude-sonnet-5)

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies placed in the compactness hierarchy

Example

Let $X$ be a set and let $\mathcal{T}_{\mathrm{disc}}$ , $\mathcal{T}_{\mathrm{ind}}$ , $\mathcal{T}_{\mathrm{cof}}$ , $\mathcal{T}_{\mathrm{coc}}$ , $\mathcal{T}_p$ and $\mathcal{T}_{\mathrm{Sier}}$ be the topologies of The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies. Compactness is as in Open cover, subcover, and compact topological space; a compact subset is a subspace that is compact in its own right. Then:

Discrete. $(X, \mathcal{T}_{\mathrm{disc}})$ is compact if and only if $X$ is finite (Finite, countably infinite, countable, uncountable).
Indiscrete. $(X, \mathcal{T}_{\mathrm{ind}})$ is compact, for every $X$ .
Cofinite. $(X, \mathcal{T}_{\mathrm{cof}})$ is compact, for every $X$ .
Particular point. For $p \in X$ , the space $(X, \mathcal{T}_p)$ is compact if and only if $X$ is finite.
Cocountable. $(\mathbb{R}, \mathcal{T}_{\mathrm{coc}})$ is neither compact nor countably compact (Countably compact, Lindel"of, sequentially compact, limit point compact and $\sigma$ -compact spaces, and relatively compact subsets).
Sierpinski. Sierpinski space $S = \{a,b\}$ is compact, being finite, and its subset $\{b\}$ is a compact subset that is not closed (In a Hausdorff space a point and a disjoint compact set, and two disjoint compact sets, have disjoint open neighbourhoods; hence every compact subset is closed, and in a compact Hausdorff space the compact subsets are exactly the closed ones).

Claims 2 and 3 are the ones worth noticing: compactness on its own is a very weak condition, and a space can be compact while separating no two of its points at all.

Facts & Assumptions

Given: A set $X$ , a point $p \in X$ where the particular-point topology is in play, and the six topologies of The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies.

[L1]

A space is compact when every family of open sets with union the space has a finite subfamily with union the space; a family is finite when it is empty or listable; every space listed as $\{x_0, \dots, x_n\}$ is compact (Open cover, subcover, and compact topological space; a compact subset is a subspace that is compact in its own right, Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison).

[L2]

$\mathcal{T}_{\mathrm{disc}} = \mathcal{P}(X)$ ; $\mathcal{T}_{\mathrm{ind}} = \{\varnothing, X\}$ ; $\mathcal{T}_{\mathrm{cof}}$ consists of $\varnothing$ and the sets with finite complement; $\mathcal{T}_{\mathrm{coc}}$ of $\varnothing$ and the sets with at most countable complement; $\mathcal{T}_p$ of $\varnothing$ and the sets containing $p$ ; and $\mathcal{T}_{\mathrm{Sier}} = \{\varnothing, \{b\}, S\}$ on $S = \{a,b\}$ (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies).

[L3]

A set is finite when it is equinumerous with a natural number, equivalently when it is listable as $\{x_0, \dots, x_n\}$ or empty; a subset of an at most countable set is at most countable; $\mathbb{R}$ is uncountable (Finite, countably infinite, countable, uncountable, Every subset of an at most countable set is at most countable, $\mathbb{R}$ is uncountable (Cantor's nested intervals, 1874)).

[L4]

A subset $A$ is a compact subset when the subspace it carries is compact, and equivalently when every family of ambient open sets covering $A$ has finitely many members covering it (Open cover, subcover, and compact topological space; a compact subset is a subspace that is compact in its own right, Subspace topology: the traces of the open sets, its closed sets and its bases, the continuity of the inclusion, and the characteristic property of a map into a subspace, A subspace is compact exactly when every family of open subsets of the ambient space covering it, indexed or not, has finitely many members covering it).

[L5]

In a Hausdorff space every compact subset is closed (In a Hausdorff space a point and a disjoint compact set, and two disjoint compact sets, have disjoint open neighbourhoods; hence every compact subset is closed, and in a compact Hausdorff space the compact subsets are exactly the closed ones, claim 3; Hausdorff space: distinct points have disjoint open neighbourhoods; every metrizable space is Hausdorff and the indiscrete topology on two points is not).

[L6]

The naturals embed in $\mathbb{R}$ by the canonical natural $\iota$ (The canonical natural $\iota(n) = n \cdot 1_F$ of a field, The natural numbers $\mathbb{N}$ (von Neumann)); $\iota$ is injective, which is a lemma about ordered fields and not part of the definition (Canonical naturals are positive and strictly increasing), so $\{\iota(n) : n \in \mathbb{N}\}$ is a countably infinite subset of $\mathbb{R}$ .

Verification

technique · direct

1.1

Claim 1. If $X$ is finite it is compact by [L1], whatever its topology. If $X$ is compact in the discrete topology, the family of singletons $\{\, \{x\} : x \in X \,\}$ is an open cover by [L2], so finitely many singletons cover $X$ and $X$ is listable, hence finite by [L3].

L1L2L3

1.2

Claim 2. Let $\mathcal{U} \subseteq \mathcal{T}_{\mathrm{ind}}$ have union $X$ . If $X = \varnothing$ the empty subfamily covers it; otherwise some member is nonempty, hence equals $X$ by [L2], and that single member is a finite subcover.

L1L2

1.3

Claim 4. If $X$ is finite it is compact by [L1]. If $X$ is compact in $\mathcal{T}_p$ , then $\{\, \{p,x\} : x \in X \,\}$ is a family of open sets by [L2] with union $X$ , so finitely many of its members cover $X$ ; their union is a listable set, so $X$ is finite by [L3].

L1L2L3

1.4

Claim 6. $S$ is finite, hence compact by [L1]; the subspace $\{b\}$ is a one-point space and so compact by [L1], making $\{b\}$ a compact subset by [L4]; and $\{b\}$ is not closed, its complement $\{a\}$ not lying in $\mathcal{T}_{\mathrm{Sier}}$ by [L2]. By [L5] this forces $S$ not to be Hausdorff, which it is not: the only open set containing $a$ is $S$ .

L1L2L4L5

2.1

Claim 3. Let $\mathcal{U} \subseteq \mathcal{T}_{\mathrm{cof}}$ have union $X \ne \varnothing$ , the empty case being as in step 1.2. Some $U_0 \in \mathcal{U}$ is nonempty, so $X \setminus U_0$ is finite by [L2], say $X \setminus U_0 = \{y_0, \dots, y_k\}$ or empty; in the second case $\{U_0\}$ covers $X$ , and in the first each $y_j$ lies in some member of $\mathcal{U}$ , and finitely many members named in this way together with $U_0$ cover $X$ .

L1L2L3

2.2

Claim 5. Put $a_n := \iota(n)$ for $n \in \mathbb{N}$ and $A_n := \{\, a_m : m > n \,\}$ , an at most countable subset of $\mathbb{R}$ by [L6] and [L3], so that $U_n := \mathbb{R} \setminus A_n$ lies in $\mathcal{T}_{\mathrm{coc}}$ by [L2]. Every real lies in some $U_n$ : a real that is no $a_m$ lies in $U_0$ , and $a_m$ lies in $U_m$ . So $\{\, U_n : n \in \mathbb{N} \,\}$ is an at most countable open cover of $\mathbb{R}$ .

L2L3L6step 1.1

3.1

The sets $U_n$ increase with $n$ , since the $A_n$ decrease, so the union of finitely many of them is a single $U_N$ , and $a_{N+1} \in A_N$ lies outside it. Hence this at most countable open cover has no finite subcover and $(\mathbb{R}, \mathcal{T}_{\mathrm{coc}})$ is neither countably compact nor compact, which is claim 5.

L1L3step 2.2∎

Remarks

Compactness alone separates nothing. The cofinite topology on an infinite set is compact and has the property that any two nonempty open sets meet, so it is as far from Hausdorff as a topology can be; the indiscrete topology is compact and has only two open sets. Every theorem on the companion page that concludes a separation property from compactness carries a Hausdorff hypothesis for exactly this reason. The purely covering conclusions there carry none: continuous images of compact spaces are compact, a continuous real function on a nonempty compact space attains its bounds, and products of compact spaces are compact, all without any separation hypothesis.

The discrete and the cocountable cases fail for different reasons. A discrete space fails compactness because its singletons already form a cover with nothing to thin; the cocountable topology on $\mathbb{R}$ fails it because countably many points can be shaved off one at a time and no finite stage removes them all. The second failure is at the countable level, which is why it kills countable compactness too.

Depends on

Used by

ℝ^* is homeomorphic to the unit circle by inverse stereographic projection, and ℕ^* is the ordinal space ω + 1 Example

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 109 results over 27 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

Compact space (Wikipedia) (standard reference, not scraped)
Cofinite topology (Wikipedia) (standard reference, not scraped)
Particular point topology (Wikipedia) (standard reference, not scraped)